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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A260056 Irregular triangle read by rows: coefficients T(n, k) of certain polynomials p(n, x) with exponents in increasing order, n >= 0 and 0 <= k <= 2*n.

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%I A260056 #12 Mar 08 2017 01:09:16
%S A260056 1,2,1,1,3,3,4,2,1,4,6,10,9,7,3,1,5,10,20,25,26,19,11,4,1,6,15,35,55,
%T A260056 71,70,56,34,16,5,1,7,21,56,105,161,196,197,160,106,55,22,6,1,8,28,84,
%U A260056 182,322,462,554,553,463,321,183,83,29,7,1,9,36,120,294,588,966,1338,1569,1570,1337,967,587,295,119,37,8,1,10,45,165,450,1002,1848,2892,3873,4477,4476,3874
%N A260056 Irregular triangle read by rows: coefficients T(n, k) of certain polynomials p(n, x) with exponents in increasing order, n >= 0 and 0 <= k <= 2*n.
%C A260056 The triangle is related to the triangle of trinomial coefficients.
%H A260056 G. C. Greubel, <a href="/A260056/b260056.txt">Table of n, a(n) for the first 100 rows, flattened</a>
%F A260056 T(n,0) = n+1, and T(n,k) = 0 for k < 0 or k > 2*n, and T(n+1,k) = T(n,k) + T(n,k-1) + T(n,k-2) for k > 0.
%F A260056 T(n,k) = Sum_{j=0..n} A027907(j,k) for 0 <= k <= 2*n.
%F A260056 T(n,k) = Sum_{j=0..k} (-1)^(k-j)*A027907(n+1,j+1) for 0 <= k <= 2*n.
%F A260056 T(n,k) = T(n,2*n-1-k) + (-1)^k for 0 <= k < 2*n.
%F A260056 p(n,x) = Sum_{k=0..2*n} T(n,k)*x^k = Sum_{k=0..n} (1+x+x^2)^k for n >= 0.
%F A260056 p(n,x) = ((1+x+x^2)^(n+1)-1)/(x+x^2), p(n,0) = p(n,-1) = n+1 for n >= 0.
%F A260056 p(n+1,x) = (1+x+x^2)*p(n,x)+1 for n >= 0.
%F A260056 Sum_{n>=0} p(n,x)*t^n = 1/((1-t)*(1-t*(1+x+x^2))).
%F A260056 T(n,2*n) = 1, and T(n,n) = A113682(n) for n >= 0.
%F A260056 T(n,n-1) = A246437(n+1), and T(n,n-1)+T(n,n) = A002426(n+1) for n > 0.
%F A260056 If d(n) is n-th antidiagonal sum of the triangle then: d(n) = A008937(n+1), and d(n+2)-d(n) = A001590(n+5) for n >= 0.
%F A260056 Conjecture: If a(n) is n-th antidiagonal alternating sum of the triangle then: a(n) = A004524(n+3).
%F A260056 Sum_{k=0..2*n} (-1)^k*T(n,k)^2 = (3^(n+1)-1)/2 for n >= 0.
%F A260056 Sum_{k=0..2*n} (-1)^k*(y*k+1)*T(n,k) = Sum{k=0..n} y*k+1 = (n+1)*(y*n+2)/2 for real y and n >= 0.
%F A260056 Conjecture of linear recurrence for column k: Sum_{m=0..k+2} (-1)^m*T(n+m,k)* binomial(k+2,m) = 0 for k >= 0 and n >= 0.
%e A260056 The irregular triangle T(n,k) begins:
%e A260056 n\k:  0   1   2    3    4    5    6    7    8    9   10  11  12  13  14  ...
%e A260056 0     1;
%e A260056 1     2   1   1;
%e A260056 2     3   3   4    2    1;
%e A260056 3     4   6  10    9    7    3    1;
%e A260056 4     5  10  20   25   26   19   11    4    1;
%e A260056 5     6  15  35   55   71   70   56   34   16    5    1;
%e A260056 6     7  21  56  105  161  196  197  160  106   55   22   6   1;
%e A260056 7     8  28  84  182  322  462  554  553  463  321  183  83  29   7   1;
%e A260056 etc.
%e A260056 The polynomial corresponding to row 2 is p(2,x) = 3+3*x+4*x^2+2*x^3+x^4.
%t A260056 A027907[n_, k_] := Sum[Binomial[n, j]*Binomial[j, k - j], {j, 0, n}]; Table[ Sum[A027907[j, k], {j, 0, n}], {n,0,10}, {k, 0, 2*n} ] // Flatten (* _G. C. Greubel_, Mar 07 2017 *)
%Y A260056 Cf. A000027 (col 0), A000217 (col 1), A000292 (col 2), A001590, A002426, A004524, A005582 (col 3), A008937, A027907, A095662 (col 5), A113682, A246437.
%K A260056 nonn,easy,tabf
%O A260056 0,2
%A A260056 _Werner Schulte_, Nov 08 2015